Question

Find the percentage error in specific resistance given by $\rho = \dfrac{{\pi {r^2}R}}{l}$ where $r$ is the radius having value $\left( {0.2 \pm 0.02} \right)\,cm$, $R$ is the resistance of $\left( {60 \pm 2} \right)\Omega$ and $l$ is the length of $\left( {150 \pm 0.1} \right)\,cm.$
A. $5.85\,\%$
B. $11.7\,\%$
C. $23.4\,\%$
D. $35.1\,\%$

Answer 1

In the question radius, resistance and the length is given. By simplifying the equation of specific resistance by using the logarithms and differentiations in the equation, we get the value of the percentage error in the specific resistance.

Complete step by step answer:
Given that $r \pm \Delta r = \left( {0.2 + 0.02} \right)\,{\text{and }}\left( {0.2 - 0.02} \right)\,cm\,$
$r \pm \Delta r = \left( {0.2 \pm 0.02} \right)\,cm$
$R \pm \Delta R = \left( {60 + 2} \right)\,\Omega \,\,{\text{and}}\left( {60 - 2} \right)\,\Omega$
$R \pm \Delta R = \left( {60 \pm 2} \right)\,\Omega$
$l \pm \Delta l = \left( {150 + 0.1} \right)\,cm\,{\text{and}}\left( {150 - 0.1} \right)$
$l \pm \Delta l = \left( {150 \pm 0.1} \right)$
Where,
$\Delta r,\Delta R,\Delta l$are the uncertainties in the radius, resistance and the length respectively, such that
$r = 0.2\,cm,\,\Delta r = 0.02\,cm$
$R = 60\,cm,\,\Delta R = 2cm$
$l = 150\,cm,\,\Delta l = 0.1\,cm$
The equation of finding the specific resistance is $\rho = \dfrac{{\pi {r^2}R}}{l}$
Applying logarithms and differentiations on the equation of specific resistance, we get
$\dfrac{{\Delta \rho }}{\rho }\, = \pm \left( {2\dfrac{{\Delta r}}{r} + \dfrac{{\Delta R}}{R} + \dfrac{{\Delta l}}{l}} \right)$
Substitute the known values in the above equation, we get
$\Rightarrow \dfrac{{\Delta \rho }}{\rho }\, = \pm \left( {2\dfrac{{0.02}}{{0.2}} + \dfrac{2}{{60}} + \dfrac{{0.1}}{{150}}} \right)$
Simplify the above equation, we get
$\Rightarrow \dfrac{{\Delta \rho }}{\rho } = 0.234$
We have to find the percentage error for the specific resistance so the values is multiple by 100
The percentage error in the specific resistance is given by,
Percentage error $\dfrac{{\Delta \rho }}{\rho } \times 100 = 0.234 \times 100$
$\Rightarrow$ Percentage error$\dfrac{{\Delta \rho }}{\rho } = 23.4\,\%$.
Therefore, the percentage error in the specific resistance is $23.4\,\%$.

Hence from the above options, option C is correct.

Note: In the question, the required parameters are given for finding the percentage error for the specific resistance. But in this case, length and change in length is not given. We have to find one of the unknowns from the formula and after that we have to find the length and change in length.